Question

A body of mass m strikes a stationary body of mass M and undergoes an elastic collision . After collision m has a speed of 1/3 rd of its initial speed. What is the ratio M/m?

Answer 1

at first calculate the final momentum of body M by laws of conservation of momentum

Answer 2

Answer:

The ratio of masses M/m = 1/2.

Explanation:

Let the velocity of mass "m" before and after collision be "u" and "v" respectively.

Let the velocity of mass "M" be "V".

Given,

v = u/3....(1) (After collision m has a speed of 1/3 rd of its initial speed)

To find,

The ratio M/m

Step (1): Conservation of momentum

momentum before collision = momentum after collision

i.e. mu = mv + MV

⇒ mu = mu/3 + MV.....(2) (from (1))

$\implies V = \frac{2}{3}\frac{mu}{M}$.....(3)

Step (2): Conservation of energy

K.E before collision = K.E after collision

$\frac{1}{2} mu^2 = \frac{1}{2} mv^2+\frac{1}{2} MV^2\\\implies \frac{1}{2} mu^2 = \frac{1}{2} m(\frac{1}{3}u ^2)+\frac{1}{2} MV^2$

Now from (3)

We get

$\frac{1}{2} mu^2 = \frac{1}{2} m(\frac{1}{3}u ^2)+\frac{1}{2} M(\frac{2}{3}\frac{mu}{M})^2\\\implies u^2 = \frac{u^2}{9} + \frac{4mu^2}{9M} \\\implies \frac{8}{9} = \frac{4m}{9M} \\\implies \frac{m}{M} = \frac{2}{1}$

therefore, the ratio of masses M/m = 1/2.

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